Question

\(\int\frac{2x^{2}\cos(x^{2})-\sin(x^{2})}{x^{2}}dx=\)

• A. \(\frac{\sin(x^{2})}{x^{2}}+c\)
• B. \(\frac{\cos(x^{2})}{x^{2}}+c\)
• C. \(\sin(x^{2})+c\)
• D. \(\frac{\sin(x^{2})}{x}+c\)

Hint:

Recognize the derivative of a quotient to simplify the integral.

Answer 1

The Correct Option is D

Step 1: Rewrite the integral.
Let I = \(\int\frac{2x^{2}\cos(x^{2})-\sin(x^{2})}{x^{2}}dx\).
We can rewrite the integral as: I = \(\int\left(2\cos(x^{2}) - \frac{\sin(x^{2})}{x^{2}}\right)dx\)

Step 2: Use substitution.
Let u = x^2. Then \(\frac{du}{dx} = 2x\), so du = 2x dx.
We can rewrite the integral as: I = \(\int\left(2\cos(u) - \frac{\sin(u)}{u}\right)dx\)

Step 3: Recognize the derivative of a quotient.
We can rewrite the integral as: I = \(\int\left(\frac{2x^2\cos(x^2) - \sin(x^2)}{x^2}\right)dx\)
Notice that this looks like the derivative of a quotient.
Let f(x) = \(\frac{\sin(x^2)}{x}\).
Then f'(x) = \(\frac{x(2x\cos(x^2)) - \sin(x^2)(1)}{x^2} = \frac{2x^2\cos(x^2) - \sin(x^2)}{x^2}\).

Step 4: Integrate.
Since f'(x) = \(\frac{2x^2\cos(x^2) - \sin(x^2)}{x^2}\), we have:
I = \(\int f'(x) dx = f(x) + c = \frac{\sin(x^2)}{x} + c\)
Therefore, the integral is \(\frac{\sin(x^{2})}{x}+c\).